Hydraulic Engineering: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

Continuity & Bernoulli

Continuity: A₁v₁ = A₂v₂
Bernoulli: p₁ + ½ρv₁² + ρgz₁ = p₂ + ½ρv₂² + ρgz₂
Flow rate: Q = A·v
Circle area: A = πd²/4

Pipe friction & head loss

Reynolds number: Re = V·D/ν = ρ·V·D/μ
Flow regime: laminar below Re ≈ 2000, turbulent above Re ≈ 4000
Mean velocity: V = Q/A, A = π·D²/4
Darcy–Weisbach: h_f = f·(L/D)·V²/(2g)
Minor losses: h_m = ΣK·V²/(2g)
Typical K values: entrance 0.5, exit 1.0, 90° bend ≈ 0.3
Pressure drop: Δp = ρ·g·h_f
Two open reservoirs: Δz = (f·L/D + ΣK)·V²/(2g)
Free outlet: H = (1 + f·L/D + ΣK)·V²/(2g)

Pumps & pump power

Hydraulic power: P_h = ρ·g·Q·H = Q·Δp
Shaft power: P_shaft = P_h/η_pump
Electrical power: P_el = P_shaft/η_motor
Pump head: H = Δz + Δp/(ρ·g) + h_L
System losses: h_L = (f·L/D + ΣK)·V²/(2g)

Open-channel flow (Manning)

Manning equation: V = (1/n)·R^2/3·S^1/2
Flow rate: Q = A·V
Hydraulic radius: R = A/P (P = wetted perimeter, free surface excluded)
Rectangle: A = b·y, P = b + 2y
Trapezoid (side slope z:1): A = (b + z·y)·y, P = b + 2y·√(1 + z²)
Circular pipe, full or half full: R = D/4
Required slope: S = (n·Q/(A·R^2/3))²

2Practice problems

1.Hydrostatic forceMedium

A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 3 mH = 1.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of glycerin, with its top edge 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 3 mH = 1 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 290 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 150 mmd₂ = 75 mmp₁ = 290 kPav₁ = 2.75 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 1.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure p_C at point C. The atmospheric pressure is 101.3 kPa.

d = 25 mmz_C = +0.5 mz₂ = −1.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2.5 m/s through a pipe 100 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2.5 m/sL = 100 mD = 200 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 7 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.

Δz = 7 mL = 100 mD = 100 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump with an efficiency of 72% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 18 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 15 L/sH = 18 mη = 72%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 23 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 75 L/sΔz = 23 mL = 150 mD = 200 mmf = 0.02ΣK = 8

H = m

9.Open-channel flow (Manning)Medium

A trapezoidal gravel channel has a bottom width of 6 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.9 m deep. Manning n = 0.025 and the bed slope is S = 0.002. Find the flow rate Q in m³/s.

b = 6 my = 1.9 mz = 1n = 0.025S = 0.002

Q = m³/s

10.Open-channel flow (Manning)Hard

A trapezoidal unfinished concrete channel with a bottom width of 5 m and side slopes of 2:1 (horizontal:vertical) must carry 8.6 m³/s in uniform flow at a depth of 0.8 m. Manning n = 0.014. Find the required bed slope S (in m/m).

Q = 8.6 m³/sb = 5 my = 0.8 mz = 2n = 0.014

S = m/m

3Solutions

1.Hydrostatic force

Answer: F = 110.4 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 3 + 1.5/2 = 3.75 m

A = H·b = 1.5·2 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.75·3 = 110.4 kN

Answer

F = 110.4 kN

2.Hydrostatic force

Answer: y_cp = 3.524 m

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 3 + 1/2 = 3.5 m

A = H·b = 1·3 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·3.5·3 = 129.8 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

I_c = b·H³/12 = 3·1³/12 = 0.25 m⁴

y_cp = y_c + I_c/(y_c·A) = 3.5 + 0.25/(3.5·3) = 3.524 m

Answer

y_cp = 3.524 m

3.Continuity & Bernoulli

Answer: p₂ = 233.3 kPa

Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 2.75·(150/75)² = 11 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.75² − 11²)/1000 = −56.72 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 290 + (−56.72) = 233.3 kPa

Answer

p₂ = 233.3 kPa

4.Continuity & Bernoulli

Answer: p_C = 81.68 kPa

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·1.5) = √(2·9.81·1.5) = 5.425 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(5.425²/2 + 9.81·0.5)/1000 = 101.3 − 19.62 = 81.68 kPa

Answer

p_C = 81.68 kPa

If this pressure falls to the vapour pressure of the water, the siphon stops working.

5.Pipe friction & head loss

Answer: h_f = 3.186 m

Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss h_f = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses h_m = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = h_f + h_m, with the exit loss K = 1 included.

Velocity head

V²/(2g) = 2.5²/(2·9.81) = 0.3186 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(100/0.2)·2.5²/(2·9.81) = 3.186 m

Answer

h_f = 3.186 m

6.Pipe friction & head loss

Answer: Q = 16.89 L/s

Loss coefficients

f·L/D = 0.025·100/0.1 = 25

ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

7 = (25 + 4.7)·V²/(2·9.81)

V = √(2·9.81·7/29.7) = 2.15 m/s

Flow rate

Q = V·π·D²/4 = 2.15·0.007854 = 0.01689 m³/s = 16.89 L/s

Answer

Q = 16.89 L/s

7.Pumps & pump power

Answer: P_shaft = 3.679 kW

Pump power: the power given to the water is the hydraulic power P_h = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: P_shaft = P_h/η_pump, and the motor draws P_el = P_shaft/η_motor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

P_h = ρ·g·Q·H = 1000·9.81·0.015·18 = 2.649 kW

Shaft power

P_shaft = P_h/η = 2.649/0.72 = 3.679 kW

Answer

P_shaft = 3.679 kW

8.Pumps & pump power

Answer: H = 29.68 m

Velocity

V = Q/A = 0.075/0.03142 = 2.387 m/s

V²/(2g) = 0.2905 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (15 + 8)·0.2905 = 6.681 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 23 + 6.681 = 29.68 m

Answer

H = 29.68 m

9.Open-channel flow (Manning)

Answer: Q = 32.3 m³/s

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R^2/3·S^1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Geometry

A = (b + z·y)·y = (6 + 1·1.9)·1.9 = 15.01 m²

P = b + 2y·√(1 + z²) = 6 + 2·1.9·√(1 + 1²) = 11.37 m

R = A/P = 1.32 m

Velocity

V = (1/0.025)·1.32^2/3·0.002^1/2 = 2.152 m/s

Flow rate

Q = A·V = 15.01·2.152 = 32.3 m³/s

Answer

Q = 32.3 m³/s

10.Open-channel flow (Manning)

Answer: S = 0.000993 m/m

Geometry

A = (b + z·y)·y = (5 + 2·0.8)·0.8 = 5.28 m²

P = b + 2y·√(1 + z²) = 5 + 2·0.8·√(1 + 2²) = 8.578 m

R = A/P = 0.6155 m

Solve Manning for S

S^1/2 = n·Q/(A·R^2/3) = 0.014·8.6/(5.28·0.7236) = 0.03151

S = 0.03151² = 0.000993

Answer

S = 0.000993

That is a fall of 0.993 m per kilometre.

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